A population of values has a normal distribution with μ=99.6 and σ=35.1. You intend to...

Step 1
From the provided information,
Mean ${\displaystyle \left(\mu \right)=99.6}$
Standard deviation ${\displaystyle \left(\sigma \right)=35.1}$
${\displaystyle X\sim N(99.6,35.1)}$
Sample size ${\displaystyle \left(n\right)=84}$
Step 2
The required probability that a sample of size ${\displaystyle n=84}$ is randomly selected with a mean between 98.5 and 100.7 can be obtained as:
and 100.7 can be obtained as:
${\displaystyle P\left(98.5<\bar{X}<100.7\right)=P\left(\frac{98.5-99.6}{\frac{35.1}{\sqrt{84}}}<\frac{x-\mu }{\frac{\sigma }{\sqrt{n}}}<\frac{100.7-99.6}{\frac{35.1}{\sqrt{84}}}\right)}$
${\displaystyle =P(-0.287<Z<0.287)}$
${\displaystyle =P\left(Z<0.287\right)-P(Z<-0.287)}$
${\displaystyle =P\left(Z<0.287\right)-[1-P\left(z<0.287\right)]}$
${\displaystyle =2P\left(Z<0.287\right)-1}$
${\displaystyle =2\left(0.6129\right)-1=0.2258}$ (Using standard normal table)
Thus, the required probability is 0.2258.